Sgu 180 greedy

I have to say that I did not understand the meaning of the question. I can only read the report to understand the question. The general idea is that there are n chains and each chain has L rings, you can remove a chain from it in one minute, and then connect the two chains. Ask the minimum time to connect these chains into one chain. Remove the shortest chain and start from the longest chain.

# Include <cstdio> # include <iostream> # include <cstring> # include <algorithm> using namespace STD; int main () {// freopen ("in.txt ", "r", stdin); int n, a [110], ANS = 0; CIN> N; For (INT I = 0; I <n; I ++) cin> A [I]; sort (A, A + n); For (INT I = 0; I <n; I ++) {if (I> 0) A [I] + = A [I-1]; if (a [I] = n-1-i-1) {ans = n-1-i-1; break;} else if (a [I]> = n-1-i) {ans = n-i-1; break; }}cout <ans <Endl; return 0 ;}